I am trying to understand the definition of a G-CV(V)-complex given by Costenoble and Warner. It seems to me that there are two different definitions.
Let $G$ be a finite group and let V be an orthogonal $G$-representation. For a subgroup $H$, let $V(H)$ denote the orthogonal complement of $V^H$ in $V$. Finally, let $|V|=\dim V$.
1.- (In Equivariant homotopy and cohomology theory) A G-CW(V) complex is a $G$-space $X$ that admits a decomposition $$X=\bigcup_{n\geq 0} X^n$$ where $X^0$ is a disjoint union of orbits of the form $G/H$ such that $H$ acts trivially on $V$ and $X^n$ is obtained from $X^{n-1}$ by attaching cells of the form $G\times_H D(V(H)\oplus \mathbb R^t)$, $t=n-|V(H)|$, along attaching $G$-maps $G\times_H S(V\oplus \mathbb R^t)$.
2.- (In Equivariant Ordinary Homology and Cohomology): As before, but now the cells are of the form $G\times_H D(V\oplus \mathbb R^t)$ with the condition that if $t<0$, then $|V^H|\geq -t$, so that $V\oplus \mathbb R^t$ is an actual $H$-representation.
Now, it seems hard to me to see why both definitions should agree. Particularly in the case where $H=\{e\}$ is the trivial group: $V(H)=\{0\}$ as $V^{\{e\}}=V$, so the disk is always going to carry the trivial action in the former definition but not in the latter.
The boundary spheres in 1.- also look weird, as they do not match what is inside the disk. Can someone please clarify this?
Now, in order to get familiar with this concept, I am trying to describe the G-CW(V) complex structure of a representation sphere $S^V$, understood as the one-point compactification of $V$. I am not sure this is indeed a G-CW(V) complex, although I think there are good reasons to expect it to be.
To keep things simple, $G=C_2$, the group with only two elements, so that $V=\mathbb R^{p,q}$, with the following conventions: $\dim R^{p,q}=p$ and the non-trivial element acts there by changing the sign of the last $q$ coordinates.
To be able to draw dome pictures, I was considering first $X=S^{2,1}$. Then $X^0=C_2$, as $\{e\}$ is the only subgroup that acts trivially on $V$. Now, for $X^1$, I was thinking that I could attach some $D^1$, as in the non-equivariant case. However, by using the first definition, no matter whether I use $\{e\}$ or $G$, I end up with a trivial action on the cell, which seems to be wrong, as $X^0$ is non-trivial and $X^1$ should ''extend'' that behaviour. I mean, I was expecting $X^1$ to be $S^{1,1}$. If, on the other hand, I use the second definition, there is a copy of $\mathbb R^{-1}$ that I do not know how to interpret. Does it make $V\oplus \mathbb R^{-1}$ into $\mathbb R^{1,1}$?